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trapecia [35]
3 years ago
15

cho hàm số y = f(x) có đạo hàm liên tục trên R thỏa mãn 2f(2x)+f(1-2x) = 12x2 víu mọi x thuộc R . Phương trình tiếp tuyến của đồ

thị hàm số y =f(x) tại điểm có hành độ x=0 là
Mathematics
1 answer:
Nina [5.8K]3 years ago
6 0

Answer:

ask in english

Step-by-step explanation:

then I can help u

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Use the point P(3, —6). Write the ordered pair for the final location of the given point after completing the transformations in
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pk oma gawd hey you

Step-by-step explanation:

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2 years ago
What is the number that is 25 units away from -12?
Crazy boy [7]

Answer:

13

Step-by-step explanation:

25 - 12 = 13

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3 years ago
25 POINTS HURRY Brainliest!!!
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Keith exercises x hours a week. Troy exercises three hours less than Keith and Joe exercises two times more than Troy. Which exp
umka21 [38]
If Keith is represented by x, and Joe is three less than Keith, Troy is represented by x-3. 
Then, we know that Joe exercises for twice the amount of time as Troy, we multiply the expression for Joe by two. 

Keith = x
Troy = x-3
Joe = 2(x-3)
4 0
3 years ago
Topic: The Quadratic Formula
Finger [1]

Answer:

Step-by-step explanation:

The quadratic formula for a equation of form

ax²+bx + c = 0 is

x= \frac{-b +- \sqrt{b^2-4ac} }{2a}

For the first equation,

x²+3x-4=0,

we can match that up with the form

ax²+bx + c = 0

to get that

ax² =  x²

divide both sides by x²

a=1

3x = bx

divide both sides by x

3 = b

-4 = c

. We can match this up because no constant multiplied by x could equal x² and no constant multiplied by another constant could equal x, so corresponding terms must match up.

Plugging our values into the equation, we get

x= \frac{-3 +- \sqrt{3^2-4(1)(-4)} }{2(1)} \\= \frac{-3+-\sqrt{25} }{2} \\ = \frac{-3+-5}{2} \\= -8/2 or 2/2\\=  -4 or 1

as our possible solutions

Plugging our values back into the equation, x²+3x-4=0, we see that both f(-4) and f(1) are equal to 0. Therefore, this has 2 real solutions.

Next, we have

x²+3x+4=0

Matching coefficients up, we can see that a = 1, b=3, and c=4. The quadratic equation is thus

x= \frac{-3 +- \sqrt{3^2-4(1)(4)} }{2(1)}\\= \frac{-3 +- \sqrt{9-16} }{2}\\= \frac{-3 +- \sqrt{-7} }{2}\\

Because √-7 is not a real number, this has no real solutions. However,

(-3 + √-7)/2 and (-3 - √-7)/2 are both possible complex solutions, so this has two complex solutions

Finally, for

4x² + 1= 4x,

we can start by subtracting 4x from both sides to maintain the desired form, resulting in

4x²-4x+1=0

Then, a=4, b=-4, and c=1, making our equation

x=\frac{-(-4) +- \sqrt{(-4)^2-4(4)(1)} }{2(4)} \\= \frac{4+-\sqrt{16-16} }{8} \\= \frac{4+-0}{8} \\= 1/2

Plugging 1/2 into 4x²+1=4x, this works as the only solution. This equation has one real solution

7 0
3 years ago
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